By Alessandra Lunardi

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**Extra info for An Introduction to Interpolation Theory**

**Sample text**

8 we know that DAi (σ, p) = {f ∈ X : s → f (x1 , . . , xi−1 , s, xi+1 , . . , xn ) ∈ W σ,p (R)}, so that n DAi (σ, p) = W σ,p (Rn ). 10 we get (X, K m )θ,p = (Lp , W k,p )θ,p = {f ∈ W k,p (Rn : Dα f ∈ W σ,p (Rn ), |α| = k} = W θm,p (Rn ). 4. 7 Let A be the realization of the Laplace operator ∆ in X = Lp (Rn ), 1 ≤ p < ∞. Then for 0 < α < 1 W α,p (Rn ) = DA (α/2, p), W α+2,p (Rn ) = DA (α/2 + 1, p). If A is the realization of ∆ in X = BU C(Rn ), in X = Cb (Rn ) or in X = L∞ (Rn ), then Cbα (Rn ) = DA (α/2, ∞), Cbα+2 (Rn ) = DA (α/2 + 1, ∞).

This is well known if X = C, and may be recovered for general X by the following argument. For every ζ ∈ Ω let x ∈ X be such that f (ζ) X = f (ζ), x and x X = 1. Applying the maximum principle to the complex function z → f (z), x we get f (ζ) X = | f (ζ), x | ≤ max{| f (z), x | : z ∈ ∂Ω} ≤ max{ f (z) X : z ∈ ∂Ω}. The maximum principle holds also for functions defined in strips. Dealing with complex interpolation, we shall consider the strip S = {z = x + iy ∈ C : 0 ≤ x ≤ 1}. If f : S → X is holomorphic in the interior of S, continuous and bounded in S, then for each ζ ∈ S f (ζ) X ≤ max{sup f (it) X , sup f (1 + it) X }.

The first one is the case m = 1. 11 For every p ∈ [1, ∞] and θ ∈ (0, 1) we have (X, K 1 )θ,p = DA (θ, p) ∩ DB (θ, p), and the norms x are equivalent. (X,K 1 )θ,p , x DA (θ,p) + x DB (θ,p) Interpolation and domains of operators 55 Proof. The embedding (X, K 1 )θ,p ⊂ DA (θ, p) ∩ DB (θ, p) is obvious, since K 1 = D(A) ∩ D(B) is continuously embedded both in D(A) and in D(B). Let x ∈ DA (θ, p) ∩ DB (θ, p). 1) that the functions λ → λθ AR(λ, A)x , λ → λθ BR(λ, B)x , λ > 0, belong to Lp∗ (0, ∞) and their norms are less than C x For every λ > 0 set DA (θ,p) , C x DB (θ,p) , respectively.